3.83 \(\int \frac{1}{x^2 (c+(a+b x)^2)} \, dx\)
Optimal. Leaf size=79 \[ -\frac{2 a b \log (x)}{\left (a^2+c\right )^2}+\frac{a b \log \left ((a+b x)^2+c\right )}{\left (a^2+c\right )^2}+\frac{b \left (a^2-c\right ) \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2+c\right )^2}-\frac{1}{x \left (a^2+c\right )} \]
[Out]
-(1/((a^2 + c)*x)) + (b*(a^2 - c)*ArcTan[(a + b*x)/Sqrt[c]])/(Sqrt[c]*(a^2 + c)^2) - (2*a*b*Log[x])/(a^2 + c)^
2 + (a*b*Log[c + (a + b*x)^2])/(a^2 + c)^2
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Rubi [A] time = 0.0861113, antiderivative size = 79, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used =
{371, 710, 801, 635, 203, 260} \[ -\frac{2 a b \log (x)}{\left (a^2+c\right )^2}+\frac{a b \log \left ((a+b x)^2+c\right )}{\left (a^2+c\right )^2}+\frac{b \left (a^2-c\right ) \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2+c\right )^2}-\frac{1}{x \left (a^2+c\right )} \]
Antiderivative was successfully verified.
[In]
Int[1/(x^2*(c + (a + b*x)^2)),x]
[Out]
-(1/((a^2 + c)*x)) + (b*(a^2 - c)*ArcTan[(a + b*x)/Sqrt[c]])/(Sqrt[c]*(a^2 + c)^2) - (2*a*b*Log[x])/(a^2 + c)^
2 + (a*b*Log[c + (a + b*x)^2])/(a^2 + c)^2
Rule 371
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Rule 710
Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]
Rule 801
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]
Rule 635
Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && !NiceSqrtQ[-(a*c)]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 260
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]
Rubi steps
\begin{align*} \int \frac{1}{x^2 \left (c+(a+b x)^2\right )} \, dx &=b \operatorname{Subst}\left (\int \frac{1}{(-a+x)^2 \left (c+x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac{1}{\left (a^2+c\right ) x}+\frac{b \operatorname{Subst}\left (\int \frac{-a-x}{(-a+x) \left (c+x^2\right )} \, dx,x,a+b x\right )}{a^2+c}\\ &=-\frac{1}{\left (a^2+c\right ) x}+\frac{b \operatorname{Subst}\left (\int \left (\frac{2 a}{\left (a^2+c\right ) (a-x)}+\frac{a^2-c+2 a x}{\left (a^2+c\right ) \left (c+x^2\right )}\right ) \, dx,x,a+b x\right )}{a^2+c}\\ &=-\frac{1}{\left (a^2+c\right ) x}-\frac{2 a b \log (x)}{\left (a^2+c\right )^2}+\frac{b \operatorname{Subst}\left (\int \frac{a^2-c+2 a x}{c+x^2} \, dx,x,a+b x\right )}{\left (a^2+c\right )^2}\\ &=-\frac{1}{\left (a^2+c\right ) x}-\frac{2 a b \log (x)}{\left (a^2+c\right )^2}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x}{c+x^2} \, dx,x,a+b x\right )}{\left (a^2+c\right )^2}+\frac{\left (b \left (a^2-c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+x^2} \, dx,x,a+b x\right )}{\left (a^2+c\right )^2}\\ &=-\frac{1}{\left (a^2+c\right ) x}+\frac{b \left (a^2-c\right ) \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2+c\right )^2}-\frac{2 a b \log (x)}{\left (a^2+c\right )^2}+\frac{a b \log \left (c+(a+b x)^2\right )}{\left (a^2+c\right )^2}\\ \end{align*}
Mathematica [A] time = 0.0437477, size = 81, normalized size = 1.03 \[ \frac{b x \left (a^2-c\right ) \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )-\sqrt{c} \left (-a b x \log \left (a^2+2 a b x+b^2 x^2+c\right )+a^2+2 a b x \log (x)+c\right )}{\sqrt{c} x \left (a^2+c\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x^2*(c + (a + b*x)^2)),x]
[Out]
(b*(a^2 - c)*x*ArcTan[(a + b*x)/Sqrt[c]] - Sqrt[c]*(a^2 + c + 2*a*b*x*Log[x] - a*b*x*Log[a^2 + c + 2*a*b*x + b
^2*x^2]))/(Sqrt[c]*(a^2 + c)^2*x)
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Maple [A] time = 0.007, size = 123, normalized size = 1.6 \begin{align*} -{\frac{1}{ \left ({a}^{2}+c \right ) x}}-2\,{\frac{ab\ln \left ( x \right ) }{ \left ({a}^{2}+c \right ) ^{2}}}+{\frac{ab\ln \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2}+c \right ) }{ \left ({a}^{2}+c \right ) ^{2}}}+{\frac{b{a}^{2}}{ \left ({a}^{2}+c \right ) ^{2}}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{b}{ \left ({a}^{2}+c \right ) ^{2}}\sqrt{c}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}{\frac{1}{\sqrt{c}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^2/(c+(b*x+a)^2),x)
[Out]
-1/(a^2+c)/x-2*a*b*ln(x)/(a^2+c)^2+b/(a^2+c)^2*a*ln(b^2*x^2+2*a*b*x+a^2+c)+b/(a^2+c)^2/c^(1/2)*arctan(1/2*(2*b
^2*x+2*a*b)/b/c^(1/2))*a^2-b/(a^2+c)^2*c^(1/2)*arctan(1/2*(2*b^2*x+2*a*b)/b/c^(1/2))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^2/(c+(b*x+a)^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.92129, size = 533, normalized size = 6.75 \begin{align*} \left [\frac{2 \, a b c x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 4 \, a b c x \log \left (x\right ) +{\left (a^{2} b - b c\right )} \sqrt{-c} x \log \left (\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 2 \,{\left (b x + a\right )} \sqrt{-c} - c}{b^{2} x^{2} + 2 \, a b x + a^{2} + c}\right ) - 2 \, a^{2} c - 2 \, c^{2}}{2 \,{\left (a^{4} c + 2 \, a^{2} c^{2} + c^{3}\right )} x}, \frac{a b c x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 2 \, a b c x \log \left (x\right ) +{\left (a^{2} b - b c\right )} \sqrt{c} x \arctan \left (\frac{b x + a}{\sqrt{c}}\right ) - a^{2} c - c^{2}}{{\left (a^{4} c + 2 \, a^{2} c^{2} + c^{3}\right )} x}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^2/(c+(b*x+a)^2),x, algorithm="fricas")
[Out]
[1/2*(2*a*b*c*x*log(b^2*x^2 + 2*a*b*x + a^2 + c) - 4*a*b*c*x*log(x) + (a^2*b - b*c)*sqrt(-c)*x*log((b^2*x^2 +
2*a*b*x + a^2 + 2*(b*x + a)*sqrt(-c) - c)/(b^2*x^2 + 2*a*b*x + a^2 + c)) - 2*a^2*c - 2*c^2)/((a^4*c + 2*a^2*c^
2 + c^3)*x), (a*b*c*x*log(b^2*x^2 + 2*a*b*x + a^2 + c) - 2*a*b*c*x*log(x) + (a^2*b - b*c)*sqrt(c)*x*arctan((b*
x + a)/sqrt(c)) - a^2*c - c^2)/((a^4*c + 2*a^2*c^2 + c^3)*x)]
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Sympy [B] time = 4.43378, size = 1620, normalized size = 20.51 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x**2/(c+(b*x+a)**2),x)
[Out]
-2*a*b*log(x + (-16*a**13*b**2*c/(a**2 + c)**4 + 48*a**11*b**2*c**2/(a**2 + c)**4 + 352*a**9*b**2*c**3/(a**2 +
c)**4 - 20*a**9*b**2*c/(a**2 + c)**2 + 608*a**7*b**2*c**4/(a**2 + c)**4 - 64*a**7*b**2*c**2/(a**2 + c)**2 + 4
32*a**5*b**2*c**5/(a**2 + c)**4 - 72*a**5*b**2*c**3/(a**2 + c)**2 + 36*a**5*b**2*c + 112*a**3*b**2*c**6/(a**2
+ c)**4 - 32*a**3*b**2*c**4/(a**2 + c)**2 - 88*a**3*b**2*c**2 - 4*a*b**2*c**5/(a**2 + c)**2 + 4*a*b**2*c**3)/(
a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3))/(a**2 + c)**2 + (a*b/(a**2 + c)**2 - b*sqrt(-c)*(
a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))*log(x + (-4*a**11*c*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*
(a**4 + 2*a**2*c + c**2)))**2 + 12*a**9*c**2*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c
+ c**2)))**2 + 10*a**8*b*c*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 88*a**
7*c**3*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 32*a**6*b*c**2*(a*b/(a*
*2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 36*a**5*b**2*c + 152*a**5*c**4*(a*b/(a**2
+ c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 36*a**4*b*c**3*(a*b/(a**2 + c)**2 - b*sq
rt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) - 88*a**3*b**2*c**2 + 108*a**3*c**5*(a*b/(a**2 + c)**2 - b*s
qrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 16*a**2*b*c**4*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 -
c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 4*a*b**2*c**3 + 28*a*c**6*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*
c*(a**4 + 2*a**2*c + c**2)))**2 + 2*b*c**5*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c +
c**2))))/(a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3)) + (a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2
- c)/(2*c*(a**4 + 2*a**2*c + c**2)))*log(x + (-4*a**11*c*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**
4 + 2*a**2*c + c**2)))**2 + 12*a**9*c**2*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c*
*2)))**2 + 10*a**8*b*c*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 88*a**7*c*
*3*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 32*a**6*b*c**2*(a*b/(a**2 +
c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 36*a**5*b**2*c + 152*a**5*c**4*(a*b/(a**2 + c
)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 36*a**4*b*c**3*(a*b/(a**2 + c)**2 + b*sqrt(-
c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) - 88*a**3*b**2*c**2 + 108*a**3*c**5*(a*b/(a**2 + c)**2 + b*sqrt(
-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 16*a**2*b*c**4*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/
(2*c*(a**4 + 2*a**2*c + c**2))) + 4*a*b**2*c**3 + 28*a*c**6*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a
**4 + 2*a**2*c + c**2)))**2 + 2*b*c**5*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2
))))/(a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3)) - 1/(x*(a**2 + c))
________________________________________________________________________________________
Giac [A] time = 1.11608, size = 158, normalized size = 2. \begin{align*} \frac{a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{a^{4} + 2 \, a^{2} c + c^{2}} - \frac{2 \, a b \log \left ({\left | x \right |}\right )}{a^{4} + 2 \, a^{2} c + c^{2}} + \frac{{\left (a^{2} b^{2} - b^{2} c\right )} \arctan \left (\frac{b x + a}{\sqrt{c}}\right )}{{\left (a^{4} + 2 \, a^{2} c + c^{2}\right )} b \sqrt{c}} - \frac{1}{{\left (a^{2} + c\right )} x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^2/(c+(b*x+a)^2),x, algorithm="giac")
[Out]
a*b*log(b^2*x^2 + 2*a*b*x + a^2 + c)/(a^4 + 2*a^2*c + c^2) - 2*a*b*log(abs(x))/(a^4 + 2*a^2*c + c^2) + (a^2*b^
2 - b^2*c)*arctan((b*x + a)/sqrt(c))/((a^4 + 2*a^2*c + c^2)*b*sqrt(c)) - 1/((a^2 + c)*x)